the square of any positive integer cannot be of the form 6q + 2 +6q+ 5 for any Integer q
Answers
Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that a = 6q + r, where 0< r< 6
a = 6q + r, where 0 ≤ r < 6
⇒ a2 = (6q + r)2 = 36q2 + r2 + 12qr [∵(a+b)2 = a2 + 2ab + b2]
⇒ a2 = 6(6q2 + 2qr) + r2 ...(i)
where,0 ≤ r < 6
Case I When r = 0, then putting r = 0 in Eq.(i), we get
a2 = 6 (6q2) = 6m
where, m = 6q2 is an integer.
Case II When r = 1, then putting r = 1 in Eq.(i), we get
a2 + 6 (6q2 + 2q) + 1 = 6m + 1
where, m = (6q2 + 2q) ais an integer.
Case III When r = 2, then putting r = 2 in Eq(i), we get
a2 = 6(6q2 + 4q) + 4 = 6m + 4
where, m = (6q2 + 4q) is an integer.
Case IV When r = 3,then putting r = 3 in Eq.(i), we get
a2 = 6(6q2 + 6q) + 9
= 6(6q2 + 6a) + 6 + 3
⇒ a2 = 6(6q2 + 6q + 1) + 3 = 6m + 3
where, m = (6q + 6q + 1) is an integer.
Case V when r = 4, then putting r = 4 in Eq.(i) we get
a2 = 6(6q2 + 8q) + 16
= 6(6q2 + 8q) + 12 + 4
⇒ a2 = 6(6q2 + 8q + 2) + 4 = 6m + 4
where, m =(6q2 + 8q + 2) is an integer.
Case VI When r = 5, then putting r = 5 in Eq.(i), we get
a2 = 6 (6q2 + 10q) + 25
= 6(6q2 + 10q) + 24 + 1
⇒ a2 = 6(6q2 + 10q + 4) + 1 = 6m + 1
where, m = (6q2 + 10q + 1) is an integer.
Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.